Filter strings that have a symbol appear certain times

user_10573715 0
Viewed 54

How to make a condition that accepts only in a loop strings that have '_' only two times.

a = ['q_s_e','s_e_v_e','s_v']# many more like these

in the above list only the fist item should used in the loop:

for i in a:
    if '_' in i:
         ...

this takes everything.

How to do it?

python string
5 Answers

You can use the count function from string:

a = ['q_s_e','s_e_v_e','s_v']
for i in a:
    if i.count('_') == 2:
      print ('hooray')

In simple words, count() method searches the substring in the given string and returns how many times the substring is present in it.

It also takes optional parameters start and end to specify the starting and ending positions in the string respectively.

The syntax of count() method is:

string.count(substring, start=..., end=...)

Another way is to filter the array of string at the beginning as well:

a = ['q_s_e','s_e_v_e','s_v']
for i in filter(lambda k: k.count('_') == 2, a):
    print ('hooray')
user_34170690

You can use str.count and iterate over a generator expression:

for i in (el for el in a if el.count('_') == 2):
    # do something
user_92095460

You need to specify that with a conditional using the count() method:

for item in list:
    if item.count('_') == 2:
        ...
user_84218250

Use regex, this will return the list you want:

filtered_list = [e for e in old_list if re.match('^[^_]*(_[^_]*){2}$', e)]

Regex ^[^_]*(_[^_]*){2}$ will make sure each of these element contains exactly 2 _

user_80462320

You can also solve this with numpy and remove your loop:

arr = np.array(['q_s_e','s_e_v_e','s_v'])
new_arr = arr[np.where(np.char.count(arr, "_") == 2)[0]]
user_27038020
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